Problem: Find all real numbers $a$ such that the equation
\[x^3 - ax^2 - 2ax + a^2 - 1 = 0\]has exactly one real solution in $x.$
Solution: Writing the equation as a quadratic in $a,$ we get
\[a^2 - (x^2 + 2x) a + (x^3 - 1) = a^2 - (x^2 + 2x) a + (x - 1)(x^2 + x + 1) = 0.\]We can then factor this as
\[(a - (x - 1))(a - (x^2 + x + 1)) = 0.\]So, one root in $x$ is $x = a + 1.$  We want the values of $a$ so that
\[x^2 + x + 1 - a = 0\]has no real root.  In other words, we want the discriminant to be negative.  This gives us $1 - 4(1 - a) < 0,$ or $a < \frac{3}{4}.$

Thus, the solution is $a \in \boxed{\left( -\infty, \frac{3}{4} \right)}.$